When I power an LED (a small one used as an indicator) with a 5VDC, it blows up, but when I use a resistor in series with the LED, it lights.

My instructor calls this resistor a current limiting resistor.

It is known that current in a series circuit remains the same.

How does the current limiting resistor work in this case, and why does the LED blow without the resistor, even though the current remains the same in both cases?

Below is the schematic of the circuit


simulate this circuit – Schematic created using CircuitLab

  • 14
    \$\begingroup\$ The current is not the same in both cases. The LED current is an exponential function of its voltage. With a resistor present, which drops voltage linearly with the current through it, when the LED tries to pull more current the resistor just drops more voltage. So the LED can't let too much current through because the resistor just drops away too much voltage when it tries. That's why the resistor limits the current. Without it, there is nothing to drop voltage and the LED has a huge current and destroys itself. \$\endgroup\$
    – jonk
    Oct 28, 2020 at 7:36
  • 3
    \$\begingroup\$ current in a series circuit remains the same ... you are misunderstanding this ...it says that the current is the same at every point of a circuit ... but the current is different in different circuits ... think of water pipes of different diameters \$\endgroup\$
    – jsotola
    Oct 28, 2020 at 7:47
  • \$\begingroup\$ This is quite radical example how a human who doesn't know how something works but he knows all words in the explanation how that something works imagines the most convenient actual meaning for the explanation. Then he wonders why that something blows or does nothing. \$\endgroup\$
    – user287001
    Oct 28, 2020 at 20:03
  • 1
    \$\begingroup\$ A circuit is like a spinning flywheel. A resistor is like your thumb, rubbing upon the wheel. Current is like the rim-speed of the wheel. So, as the metal wheel passes under your thumb, does it change from "high energy metal" into "low energy metal?" Nope. Instead, the entire wheel simply slows down. Even the distant parts of the flywheel are slowed just as much. Yet your thumb never touched them. How can this happen?!! \$\endgroup\$
    – wbeaty
    Oct 29, 2020 at 1:10
  • 5
    \$\begingroup\$ @Saqlain No, jonk is saying that the current through the LED will be different when the resistor is there, compared to when the resistor is not there. \$\endgroup\$
    – Hearth
    Oct 30, 2020 at 13:10

10 Answers 10


It is known that current in a series circuit remains the same.

The current is the same at any part of a simple series circuit. But that doesn't mean the current stays the same if you change the circuit.

An LED drops around 1.5 to 3V when it's working normally. If you connect it straight across a 5V supply, an excessively large current flows, and the LED blows.

Put a 100 ohm resistor in, and the extra voltage is dropped across the resistor. This will be somewhere between 2 and 3.5V. Given I = V/R, a 100 ohm resistor will pass between 0.02 and 0.035A (or 20 to 35mA). That's much better for an LED.

  • \$\begingroup\$ So, what does it mean that the current is the same at any part of a simple circuit? \$\endgroup\$
    – Saqlain
    Oct 28, 2020 at 17:54
  • 2
    \$\begingroup\$ @Saqlain In a "simple circuit" (i.e. where everything is in series with each other), the current flow through each component is the same. So when you have a LED and resistor in series, the current flow though the resistor is X amps and the current flow through the LED is also X amps. \$\endgroup\$
    – Jay
    Oct 28, 2020 at 18:51

Another question about LEDs and resistors...
Ok, I'll take it.

I believe you might benefit from the visual aid associated with the use of load lines. Try to see your circuit as the combination of two one-port elements: the generator and the load. Each of these one-ports will have its own Voltage-Current characteristic and when you combine them together,by connecting the respective ports, you end up in sharing both voltage and current.
Note that it is customary to adopt two antithetical conventions for the sign of the current: in the case of the source, the current is exiting the one-port from the point at higher potential, whereas for the load the current is entering the one-port from the point at higher potential (where the + is or the voltage arrow has its point in the non-German convention).

source and load characteristics

Now, since both one-ports share the same voltage V and the same current I (here the choice of current convention shows its usefulness), the operating point of the circuit falls on the intersection of the two characteristics curves.

operating point as intersection of source and load characteristics

You can ascribe the description each port V-I characteristic to the application of a generalized form of Ohm's law (allowing for nonlinear and active elements), while the relationship between the electric variables of the ports established by their interconnection can be seen as an application of Kirchhoff voltage and current laws.

Let's see what this entails by analyzing some simple circuits where we connect a diode to different kinds of sources. In all the following circuits, the load will be a diode (an LED, if you will) which is basically a nonlinear resistor with an exponential V-I characteristic.

Let's start with an ideal voltage source. The V-I characteristic is a vertical line at \$V_{batt}\$, meaning the generator is capable of providing the same exact voltage \$V_{batt}\$ no matter the current drawn by the load. If the value of \$V_{batt}\$ is sufficiently higher than the 'nominal' threshold voltage of the diode, the intersection of the generator and load characteristics will be at the point (Vq, Iq) where the current Iq is high enough to destroy your diode.

Ideal Voltage Source

Ideal voltage sources do not exist in real life, and any source you will find will exhibit an internal resistance \$R_s\$ that will cause the V-I characteristic to bend towards the I axis. The intersection of the source characteristic with the I axis is the short circuit current \$I_{sc}\$ (not shown in the following picture because it's way up on the I axis). In some cases the combination of the open circuit voltage \$V_{batt}\$ of the source and its internal resistance \$R_s\$ might cause the intersection of source and load curve to happen at a point (Vq, Iq) with low enough current not to destroy the LED.

Internal resistance might be enough

This is the case for example of those silverish button cells: you can safely attach an LED to them without any additional limiting resistor. But with most other voltage source - unless they are regulated by other means around the nominal threshold voltage of the LED, you will usually end up killing the diode.

So, why not take control of the situation and use a series resistor \$R_{lim}\$ in order to limit the value of the current to a safe value? This is the easiest way to power a LED (a better way would be to use a current source). Usually this series limiting resistance \$R_{lim}\$ is much higher than the internal resistance \$R_s\$ of the source so that the latter parameter can be safely ignored. The 'limited' source characteristic would bend even further toward the I axis and the intersection of source and load lines can be made to happen at a predetermined point (Vq, Iq).

Limiting series resistance

Given the shape of the exponential diode characteristic (here not drawn to scale), the value Vq won't stray too much from the nominal threshold voltage \$V_{\gamma}\$ (or \$V_{threshold}\$) of the LED, but the important thing is that by adding a series resistance \$R_{lim}\$ we gained control on the otherwise wildly variable value of the current.
The mathematics to determine the exact value of the intersection of the linear load line with the exponential diode curve, while still being basic calculus, is not exactly elementary (you might be required to add the Lambert W function to your vocabulary) and it is customary to resort to numerical or graphical approximations as shown in some of the other answers.
(These approximations amount to swap the nonlinear exponential curve with a piece-wise linear curve either rising vertically at \$V_{\gamma}\$ or slanted with a slope determined by the diode's differential resistance \$r_d\$)

From a qualitative point of view, though, you can appreciate the role played by the limiting resistor by drawing the V-I characteristics of the source for different values of \$R_{lim}\$:

Variable limiting resistor

As you can see, by increasing the value of \$R_{lim}\$ you bring down the operating point(Vq, Iq) at lower and lower values of the current. Note that for a given, set, fixed value of \$R_{lim}\$ the current is the same for all series elements (source, limiting resistor and LED) but if you change \$R_{lim}\$ (by either swapping the resistor, or acting on the knob of a potentiometer, or even by changing its temperature by blowing hot air over it) you will see the current change in all elements of the series.

This is what the application of the load lines method looks like when the exponential diode characteristic is approximated by a piecewise linear characteristic with fixed \$V_{thereshold}\$ and zero differential resistence:

approximation LED with fixed Vf

  • 3
    \$\begingroup\$ This answer is great, the hand drawings and amount of information in this is priceless. thank you. \$\endgroup\$ Nov 2, 2020 at 11:56
  • \$\begingroup\$ Noting your "Do ya feel lucky punk?" - you may enjoy my 2016 answer here :-) \$\endgroup\$
    – Russell McMahon
    Nov 4, 2020 at 8:41
  • \$\begingroup\$ @RussellMcMahon for I moment I thought I had the quote wrong. But no, it seems it was you who took a poetic license. :-) \$\endgroup\$ Nov 7, 2020 at 14:19

The short answer is, no. Current is the same throughout in a series circuit. The resistor in series changes the current, however. You knew that; and we'll come back to that later, below.

The tl; dr answer is, we use the LED's forward-voltage characteristic to set its current with the dropping resistor.

Let's talk about that.

LED Forward Voltage

The LED is a semiconductor device (a diode), and like all diodes has a conduction threshold voltage that is more-or-less constant. This is the forward voltage, or Vf.

Because of this Vf theshold, the way the LED response to applied voltage is highly non-linear:

  • below Vf, LED current will be next to nothing;
  • above Vf the current rises rapidly.

Keep going much further above Vf and the LED will burn up. This is why your LED died when connected directly to 5V.

This in mind, we know that we need to regulate the LED current to keep it in its operating sweet spot: making light, but not torching itself in the process.

But let's talk about that threshold voltage Vf for a bit.

What Determines Vf?

The Vf threshold depends on the color of the LED and its exact material formulation. In general, the shorter the emitted wavelength, the higher the LED Vf threshold.

LED I-V Characteristics For Different Colors

From here: http://lednique.com/test-equipment/testing-unknown-leds/

The datasheet for the specific LED you are using will include this I-V curve data. This, in turn, helps solve the rest of the current-setting dilemma.

How The Dropping Resistor Sets LED Current

We know that we need to regulate the LED current to get the brightness we want, yet not burn the LED in the process.

The simplest way to do that is to use a fixed supply (like 5V you've shown) and a series resistor. We take advantage of the LED’s known voltage drop (its threshold voltage Vf), and that fixed supply (5V), to set the LED current as follows:

  • I(led) = (5V - Vf) / R1

Simple, right?

Not quite. There’s a few things going on here that allow us to make that equation.

Kirchhoff Current and Voltage

The series resistor and LED currents are the same based on Kirchhoff’s Current Law for series circuits, which tells us that current in a loop circuit is the same at any point in the loop.

So we know:

  • I(led) = I(resistor)

Likewise, Kirchhoff’s Voltage Law for loop circuits says that the sum of all the voltage drops must equal the supply voltage.

And so we also know:

  • 5V = V(resistor) + Vf


  • V(resistor) = 5V - Vf

Finally, Ohm’s Law and some algebraic substitution tell us the resistor and LED current.

That is, I = E/R, so we have:

  • I(resistor) = V(resistor) / R1

After substituting (5V - Vf) for V(resistor), we are left with:

  • I(resistor) = (5V - Vf) / R1

And because of Kirchhoff's Current Law we know:

  • I(led) = I(resistor) = (5V - Vf) / R1

How To Pick The Right Dropping Resistor

We know the supply (5V), and we know Vf (from the LED data sheet, about 2-3V depending on color), and we know the current we want (about 10-20mA), so we can solve for R1:

  • R1 = (5V - Vf) / I(led)

Example: White LED with Vf of 3V, driven at 20mA:

  • R1 = (5V - 3V) / 20mA = 100 Ohms

Assume a 5mm LED @ 20 mA is about 10 to 15 Ohms for Red to White @ 2 to 3V respectively, so it is relatively small. but at very low currents this LED effective series resistance rises sharply as the diode voltage drops slowly.

Thus the total current depends on the voltage remained that drops across the series R. (5-3V) /100R = 20 mA



A diode conducts current if it is in forward bias. The diode is forward biased when the voltage drop across it is equal to it's forward voltage. If the diode is forward biased, then the voltage across won't go any higher.

A silicon diode usually has a forward voltage of \$V_{\text{fwd}}=0.7 \text{V} \$. Let's look at your circuit without the resistor, and assume that the diode is a silicon diode.


simulate this circuit – Schematic created using CircuitLab

From inspection the diode is in forward bias (if this is not clear I can elaborate), meaning that the voltage across the diode is \$0.7 \text{V} \$.

Because our voltage source provides \$5 \text{V} \$ there must be a voltage drop of \$4.3 \text{V} \$ across the resistance in the wire, due to KVL.

A wire has very low resistance (in the magnitude of micro-ohms). Let's set the resistance of the wire to \$0.1 \text{m}\Omega \$. Using Ohm's Law the current is \$I=\frac{V}{R}=\frac{4.3 \text{V}}{0.1 \text{m}\Omega}= 43000 \text{A}\$ (!!!!!!!)

Due to KCL, this current also runs through the diode and the voltage source.

Such ridiculously high current will obviously destroy your components. We don't want this to happen, so we introduce series resistance to lower the current.


simulate this circuit

There is still a voltage drop across of \$4.3 \text{V} \$, but it is now across a much greater resistance, than the resistance of the wire. Using Ohm's Law now the current is \$ I=\frac{V}{R}=\frac{4.3 \text{V}}{100\Omega}=0.043 \text{A} = 43 \text{mA}\$

Due to KCL, this current also flows through the diode and the voltage source.

This current is significantly lower than before and, depending on the ratings of the diode, won't damage any components in the circuit.


The LED cannot withstand too much current, and that is why you need the resistor. The consumed power is P = V*I and since I =V/R, means that P= V^2/R. Now when you only have the LED, the LEDs DC resistance is quite low and then you get a lot of current flowing through it. When you add the resistor, the total resistance seen by the battery is R + Rled, so that reduces the power flow since P=V^2/R

  • \$\begingroup\$ P = V^2/R doesn’t hold for a diode, because a diode doesn’t follow Ohm’s Law, V = RI. \$\endgroup\$
    – Carl
    Nov 6, 2020 at 15:06

You've probably heard this old adage before: It's not the voltage, but the current that'll kill you! when talking about electrocution. Contact with a high voltage on a fairly low resistance (skin), will generate a current. The physical electrons passing through an object is what destroys it. Yes, a higher voltage generates a higher current, but the phenomenon of current is what is destructive.

Now let's go to your LED circuit. The LED burns out in case #1 because the current through it is too high. If you constructed this circuit, you actually still have a resistor; it's the wiring. For case #2, your resistor limits the current to be below the maximum rating of the LED. If you want to calculate and check what the currents are, follow these steps.

  1. The voltage across the LED is called the "Forward voltage drop". You can usually look this up.
  2. The voltage across R1 is whatever's leftover. So take your voltage source and subtract the forward voltage drop. You can actually measure this with a voltmeter, just measure across the 2 leads of the resistor.
  3. The current through the circuit is VR1 / R1 The difference between case #1 and #2 is that R1 = ~.1ohms in case #1 and 100ohms in case #2. Using a typical LED forward voltage drop of about 2V, that means the LED was exposed to (5-3)/.1 = 20A. It won't really be that, because your power source is probably too weak, but you get the idea.

Answering the question in the title: no. The resistor does not change the current passing through it. The presence of a resistor determines the current thought the it, but there is no difference in the current between its terminals.

To understand it better, imagine there was a switch bypassing the resistor in your circuit. Start with the switch on, and you will have a very high current. When you switch it off, you force the current through the resistor. The information that the "easy" path is blocked and the only path remaining is the resistor propagates through the circuit at the speed of light (like a pressure wave propagates at the speed of sound), and eventually the current though the entire circuit will stabilize at the new value, accounting for the presence of the resistor.


In a series circuit you can always look at the voltage drop across each component. With the resistor in place, the diode only has ~1.5V and the resistor has ~3.5V. Without the resistor, the diode has all 5V across it! The LED is not rated for this and a huge amount of current will flow through the LED to try to maintain 5V across it.

The resistor is current limiting in that it reduces the amount of current from the first circuit. The current is not the same between the two circuits, and the voltage drop is not either.


When I power an LED (a small one used as an indicator) with a 5VDC, it blows up...

The only way to really understand the "odd" LED behavior which eventually makes it burn, is to realize that it behaves as a dynamic (self-variable) resistor. In contrast to the ordinary static (constant, ohmic, linear) resistor, it decreases its "resistance" when the voltage across it increases. As a result, the current vigorously increases and finally, the LED burns.

You can see this in Ohm's law I = Vinc/Rdec where both numerator and denominator change but in opposite directions (Vinc increases and Rdec decreases). As a result, their ratio enormously increases.

The power of this intuitive explanation is that you can make a simple experiment to emulate the LED. Take a variable resistor (rheostat) and connect a voltmeter in parallel. Your task is to keep the voltage across it constant and equal to the desired LED forward voltage. It is preferable to supply the rheostat through a resistor.

If you see the voltage increases, decrease the resistance to restore the voltage and v.v. You can explain this again by the help of Ohm's law but written in its dual form V = Iinc.Rdec. Now both factors change but in opposite directions (Iinc increases and Rdec decreases). As a result, their product V does not change.

Diode as a dynamic resistor

In this way, you can emulate various diodes (Si, Ge, Zener, etc.). Later you can use this setup to emulate transistors... and even negative resistors... but this is another game...


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